Sequential spaces, V

In the previous post Sequential spaces, IV, we show that the uncountable product of sequential spaces is not sequential (e.g. the product 2^{\omega_1} is not sequential). What is more remarkable is that the product of two sequential spaces needs not be sequential. We present an example of a first countable space and a Frechet space whose product is not a k-space (thus not sequential). For the previous discussion on this blog on sequential spaces and k-spaces, see the links at the end of this post.

Let \mathbb{R} be the real line and let \mathbb{N} be the set of all positive integers. Let X be the space \mathbb{R}-\left\{1,\frac{1}{2},\frac{1}{3},\cdots\right\} with the topology inherited from the usual topology on the real line. Let Y=\mathbb{R} with the positive integers identified as one point (call this point p). We claim that X \times Y is not a k-space and thus not a sequential space. To this end, we define a non-closed A \subset X \times Y such that K \cap A is closed in K for all compact K \subset X \times Y.

Let A=\bigcup \limits_{i=1}^\infty A_i where for each i \in \mathbb{N}, the set A_i is defined by the following:

    \displaystyle A_i =\left\{\biggl(\frac{1}{i}+\frac{a_i}{j},i+\frac{0.5}{j} \biggr) \in X \times Y:j \in \mathbb{N}\right\}

where \displaystyle  a_i=\biggl(\frac{1}{i}-\frac{1}{i+1} \biggr) 10^{-i}.

Clearly A is not closed as (0,p) \in \overline{A}-A. In fact in the product space X \times Y, the point (0,p) is the only limit point of the set A. Another observation is that for each n \in \mathbb{N}, (0,p) is not a limit point of \bigcup \limits_{i=1}^n A_i. Furthermore, if z_i \in A_i for each i \in S where S is an infinite subset of \mathbb{N}, then (0,p) is not a limit point of \left\{z_i:i \in S\right\}. It follows that no infinite subset of A is compact. Consequently, K \cap A is finite for each compact K \subset X \times Y. Thus X \times Y is not a k-space. To see that X \times Y is not sequential directly, observe that A is sequentially closed.

Previous posts on sequential spaces and k-spaces

2 thoughts on “Sequential spaces, V

    • Yes, “the square of a sequential space need not be sequential” (see note in Franklin, “Spaces in which sequences suffice”, 1965, Fundamenta Mathematicae, 57(1), p. 112).

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s